I-n-dx-cos-n-x-Prove-that-I-n-n-2-n-1-I-n-2-sin-x-n-1-cos-n-1-x-

Question Number 167666 by LEKOUMA last updated on 22/Mar/22

I_{n} = \int \frac{d x}{\cos^{n} x}

P r o v e t h a t

I_{n} = \frac{n - 2}{n - 1} I_{n - 2} + \frac{\sin x}{(n - 1) \cos^{n - 1} x}

Commented by peter frank last updated on 22/Mar/22

Reduction formular

Answered by chhaythean last updated on 22/Mar/22

Solution

I_{n} = \int \frac{dx}{\cos^{n} x} = \int \sec^{n} xdx

= \int \sec^{n - 2} {xsec}^{2} xdx

let {\begin{cases} u = \sec^{n - 2} x \Rightarrow du = (n - 2) \sec^{n - 2} xtanxdx \\ dv = \sec^{2} xdx \Rightarrow v = tanx \end{cases}

I_{n} = \sec^{n - 2} xtanx - (n - 2) \int \sec^{n - 2} {xtan}^{2} xdx

= \sec^{n - 2} xtanx - (n - 2) \int \sec^{n} xdx + (n - 2) \int \sec^{n - 2} dx

I_{n} = \sec^{n - 2} xtanx - (n - 2) I_{n} + (n - 2) I_{n - 2}

I_{n} + (n - 2) I_{n} = \sec^{n - 2} xtanx + (n - 2) I_{n - 2}

(n - 1) I_{n} = \sec^{n - 2} xtanx + (n - 2) I_{n - 2}

I_{n} = \frac{\frac{1}{\cos^{n - 2} x} \times \frac{sinx}{cosx}}{n - 1} + \frac{n - 2}{n - 1} I_{n - 2}

I_{n} = \frac{n - 2}{n - 1} I_{n - 2} + \frac{sinx}{\cos^{n - 1} x (n - 1)} true

So \begin{matrix} I_{n} = \frac{n - 2}{n - 1} I_{n - 2} + \frac{sinx}{(n - 1) \cos^{n - 1} x} is proved . \end{matrix}

Commented by LEKOUMA last updated on 22/Mar/22

T h a n k s

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